给定一个整数N和一个 (N – 1) 个整数的arr1[] ,任务是找到[1, N]范围内 N 个整数的序列arr2[]使得arr1[i] = arr2[i+ 1] – arr2[i] 。序列 arr1[] 中的整数位于[-N, N]范围内。
例子:
Input: N = 3, arr1[] = {-2, 1}
Output: arr2[] = {3, 1, 2}
Explanation:
arr2[1] – arr2[0] = (1 – 3) = -2 = arr1[0]
arr2[2] – arr2[1] = (2 – 1) = 1 = arr1[1]
Input: N = 5, arr1 = {1, 1, 1, 1, 1}
Output: arr2 = {1, 2, 3, 4, 5}
Explanation:
arr2[1] – arr2[0] = (2 – 1) = 1 = arr1[0]
arr2[2] – arr2[1] = (3 – 2) = 1 = arr1[1]
arr2[3] – arr2[2] = (4 – 3) = 1 = arr1[2]
arr2[4] – arr2[3] = (5 – 4) = 1 = arr1[3]
方法:
请按照以下步骤解决问题:
- 假设arr2[]的第一个元素是X 。
- 下一个元素将是X + arr1[0] 。
- 可以表示arr2[]的其余元素,wrt X 。
- 众所周知,序列arr2[]可以包含[1, N]范围内的整数。所以最小可能的整数是1 。
- 可以根据X找出arr2[]的最小数量,并将其与1相等以找到X的值。
- 最后使用X的值,可以找出arr2[] 中的所有其他数字。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the sequence
void find_seq(int arr[],
int m, int n) {
int b[n];
int x = 0;
// initializing 1st element
b[0] = x;
// Creating sequence in
// terms of x
for (int i = 0;
i < n - 1; i++) {
b[i + 1] = x +
arr[i] + b[i];
}
int mn = n;
// Finding min element
for (int i = 0; i < n; i++)
{
mn = min(mn, b[i]);
}
// Finding value of x
x = 1 - mn;
// Creating original sequence
for (int i = 0; i < n; i++) {
b[i] += x;
}
// Output original sequence
for (int i = 0; i < n; i++) {
cout << b[i] << " ";
}
cout << endl;
}
// Driver function
int main()
{
int N = 3;
int arr[] = { -2, 1 };
int M = sizeof(arr) / sizeof(int);
find_seq(arr, M, N);
return 0;
}
Java
// Java implementation of the above approach
class GFG{
// Function to find the sequence
static void find_seq(int arr[], int m,
int n)
{
int b[] = new int[n];
int x = 0;
// Initializing 1st element
b[0] = x;
// Creating sequence in
// terms of x
for(int i = 0; i < n - 1; i++)
{
b[i + 1] = x + arr[i] + b[i];
}
int mn = n;
// Finding min element
for(int i = 0; i < n; i++)
{
mn = Math.min(mn, b[i]);
}
// Finding value of x
x = 1 - mn;
// Creating original sequence
for(int i = 0; i < n; i++)
{
b[i] += x;
}
// Output original sequence
for(int i = 0; i < n; i++)
{
System.out.print(b[i] + " ");
}
System.out.println();
}
// Driver code
public static void main (String[] args)
{
int N = 3;
int arr[] = new int[]{ -2, 1 };
int M = arr.length;
find_seq(arr, M, N);
}
}
// This code is contributed by Pratima Pandey
Python3
# Python3 program for the above approach
# Function to find the sequence
def find_seq(arr, m, n):
b = []
x = 0
# Initializing 1st element
b.append(x)
# Creating sequence in
# terms of x
for i in range(n - 1):
b.append(x + arr[i] + b[i])
mn = n
# Finding min element
for i in range(n):
mn = min(mn, b[i])
# Finding value of x
x = 1 - mn
# Creating original sequence
for i in range(n):
b[i] += x
# Output original sequence
for i in range(n):
print(b[i], end = ' ')
print()
# Driver code
if __name__=='__main__':
N = 3
arr = [ -2, 1 ]
M = len(arr)
find_seq(arr, M, N)
# This code is contributed by rutvik_56
C#
// C# implementation of the above approach
using System;
class GFG{
// Function to find the sequence
static void find_seq(int []arr, int m,
int n)
{
int []b = new int[n];
int x = 0;
// Initializing 1st element
b[0] = x;
// Creating sequence in
// terms of x
for(int i = 0; i < n - 1; i++)
{
b[i + 1] = x + arr[i] + b[i];
}
int mn = n;
// Finding min element
for(int i = 0; i < n; i++)
{
mn = Math.Min(mn, b[i]);
}
// Finding value of x
x = 1 - mn;
// Creating original sequence
for(int i = 0; i < n; i++)
{
b[i] += x;
}
// Output original sequence
for(int i = 0; i < n; i++)
{
Console.Write(b[i] + " ");
}
Console.WriteLine();
}
// Driver code
public static void Main(String[] args)
{
int N = 3;
int []arr = new int[]{ -2, 1 };
int M = arr.Length;
find_seq(arr, M, N);
}
}
// This code is contributed by 29AjayKumar
Javascript
输出:
3 1 2
时间复杂度: O(N)
辅助空间: O(1)
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